Convergent and Divergent Series Examples
Lets learn first what is the convergent and divergent series.
What is convergent series and divergent series ?
A series which have finite sum is called convergent series.Otherwise is called divergent series.
$ \lim _{n \rightarrow \infty} S_{n}=S $
- If the partial sums Sn of an infinite series tend to a limit S, the series is called
convergent.
Otherwise it is called divergent. - The limiting value S is called the sum of the series
Lets look at some examples of convergent and divergence series examples.
Let us consider two series
Convergent series example
The sum of this series is finite.
$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots $
Divergent series example
The sum of this series is not finite.
$ 1+2+4+8+16+\cdots $
Tests for convergence of series
It can be difficult to know if a series is convergent or divergent.
But here is some methods that can be used to determine if the series is convergent or divergent
- Preliminary test
- Comparison test
- Ratio test
Preliminary Test
A necessary but not sufficient condition for a series of real positive terms $ \sum u_{n} $to be convergent is that the term untends to zero as n tends to infinity ie.
$ \lim _{n \rightarrow \infty} u_{n}=0 $
If this condition does not satisfy then series must diverge.But if this condition get satisfied then series can be divergent or convergent because this is not a sufficient condition for convergence.
Comparison test
In comparison test we compare our series with a series whose convergence is already known to us.
Let us consider two series $ \sum u_{n} $ and $ \sum u_{n} $ and suppose that we know the latter to be convergent.
Then, if each term un in the first series is equal to less than the each term in second series vnfor all n greaterthan some fixed number N that will vary from series to series,
then the original series $ \sum u_{n} $ is also convergent.
In other words if series $\sum v_{n} $ is convergent and,
$ u_{n} \leq v_{n} \quad $ for $ n>N $ then, $ \sum u_{n} $ converges.
However,if $ \sum v_{n} $ diverges and $ u_{n} \geq v_{n} $ for all $ n $ greater than some fixed number then $ \sum u_{n} $ diverges.
Example #1
Determine whether following series is divergent or convergent.
$ \sum_{n=1}^{\infty} \frac{1}{n !+1}$$=\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{25}+\cdots $
We compare this series with the series of $ \sum_{n=0}^{\infty} \frac{1}{n !} $
$ \sum_{n=0}^{\infty} \frac{1}{n !}=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots=2+\frac{1}{2 !}+\frac{1}{3 !}+\cdots $
Here this series is perfectly equals to expension of $e^{x}$ when x=1 i.e
$e^{1}=e=$$ 1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots $
Now we know that the series of e is convergent.Because it has a finite sum.
Since each term in the first series is less than the corresponding term in second series.
So first series is also convergent.
Ratio test
The ratio test determine whether a series converges by comparing the relative magnitudes of successive terms.
If we consider a series $\sum u_{n}$ then ratio of successive terms is,
$ \rho=\lim _{n \rightarrow \infty}\left(\frac{u_{n+1}}{u_{n}}\right) $
Convergent : if $ \rho \lt 1 $
Divergent : if $ \rho \gt 1 $
Undetermined : if $\rho = 1$
Example #2
Determine whether following series is divergent or convergent.
$ \sum_{n=0}^{\infty} \frac{1}{n !}=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots$$=2+\frac{1}{2 !}+\frac{1}{3 !}+\cdots $
By using ratio test we have,
$ \rho=\lim _{n \rightarrow \infty}\left[\frac{n !}{(n+1) !}\right]$$=\lim _{n \rightarrow \infty}\left(\frac{1}{n+1}\right)=0 $
Since $\rho \lt 1 $ so series converges.